What is the transformation law for tensor components in differential geometry?

[shooride]

I read in many books the metric tensor is rank (0,2), its inverse is (2,0) and has some property such as
$g^{\mu\nu}g_{\nu\sigma}=\delta^\mu_\sigma$ etc. My question is: what does $g^\mu_\nu$ mean?! This tensor really confuses me! At first, I simply thought that $g^{\mu\nu}\delta_{\mu\sigma}=g^\nu_\sigma$, but I realized it is not true. Is $g^\mu_\nu$ a metric, again? I mean can I write something like $g^\mu_\nu x^\nu=x^\mu$?

 

[Mentz114]

Please fix your Latex by replacing the '$' with a pair of '#'. Like this

$g^{\mu\nu}g_{\nu\sigma}=\delta^\mu_\sigma$

 

[Orodruin]

I have used my magic Mentor powers to fix the LaTeX in your original post.

I read in many books the metric tensor is rank (0,2), its inverse is (2,0) and has some property such as
$g^{\mu\nu}g_{\nu\sigma}=\delta^\mu_\sigma$ etc. My question is: what does $g^\mu_\nu$ mean?! This tensor really confuses me! At first, I simply thought that $g^{\mu\nu}\delta_{\mu\sigma}=g^\nu_\sigma$, but I realized it is not true. Is $g^\mu_\nu$ a metric, again? I mean can I write something like $g^\mu_\nu x^\nu=x^\mu$?

What do you mean by $\delta_{\mu\nu}$? The Kronecker delta tensor is a (1,1) tensor and a priori has one covariant and one contravariant index. Generally, you would not talk about $g^\mu_\nu$, by definition $g^{\mu\sigma}g_{\sigma\nu} = \delta^\mu_\nu$.

 

[shooride]

I have used my magic Mentor powers to fix the LaTeX in your original post.
What do you mean by $\delta_{\mu\nu}$? The Kronecker delta tensor is a (1,1) tensor and a priori has one covariant and one contravariant index. Generally, you would not talk about $g^\mu_\nu$, by definition $g^{\mu\sigma}g_{\sigma\nu} = \delta^\mu_\nu$.

Thanks for LaTeX. I had some problems with that.. I know that the Kronecker delta tensor is $\delta^\mu_\nu$, but we can consider $\delta_{\mu\nu}$ as the identity matrix, can't we? Anyway, my question is what does $g^\mu_\nu$ mean? How one can define it on a Riemannian manifold? or just we can't define it?!

 

[DrGreg]

For any type-2 tensor \textbf{A}<br /> A^\mu{}_\nu = g^{\mu\sigma}A_{\sigma\nu}<br />Now put \textbf{A} = \textbf{g} and what do you get?

 

[shooride]

For any type-2 tensor \textbf{A}<br /> A^\mu{}_\nu = g^{\mu\sigma}A_{\sigma\nu}<br />Now put \textbf{A} = \textbf{g} and what do you get?

Yes, the answer is $\delta^\mu{}_\nu$. So, one can simply write $g_{\mu\nu}\delta^{\nu\sigma}=g_{\mu\sigma}$, right? By the way, can one define a new metric (and not new coordinate system) such that $g'_{\mu\nu}=A_\mu{}^\sigma g_{\sigma\nu}$ where ${\bf A}$ is an arbitrary position independent tensor which and not necessarily $\delta_{\mu}{}^\nu$?!

 

[Orodruin]

as the identity matrix, can't we?

No, definitely not. Even if it has the correct components in one coordinate system, it will not in another one. If you want any kind of meaning for $g^\mu_\nu$ you must define it to be equal to $\delta^\mu_\nu$.

 

[Orodruin]

So, one can simply write $g_{\mu\nu}\delta^{\nu\sigma}=g_{\mu\sigma}$, right?

Big no no, the indices do not even match in that expression.

 

[Nugatory]

By the way, can one define a new metric (and not new coordinate system) such that $g'_{\mu\nu}=A_\mu{}^\sigma g_{\sigma\nu}$ where ${\bf A}$ is an arbitrary position independent tensor which and not necessarily $\delta_{\mu}{}^\nu$?!

No. You may get a new tensor that way, but there's no particular reason to expect it to be the metric for some manifold.

 

[shooride]

No, definitely not. Even if it has the correct components in one coordinate system, it will not in another one. If you want any kind of meaning for $g^\mu_\nu$ you must define it to be equal to $\delta^\mu_\nu$.

O.K, I understand $\delta_{\mu\nu}$ is not a tensor, but can't we define $\delta_{\mu\nu}$? For example, can't we write $\partial_\mu \partial_\nu x^2=2\delta_{\mu\nu}$? I know it is not a definition..but I think that we use $\delta_{\mu\nu}$..

 

[Orodruin]

O.K, I understand $\delta_{\mu\nu}$ is not a tensor, but can't we define $\delta_{\mu\nu}$? For example, can't we write $\partial_\mu \partial_\nu x^2=2\delta_{\mu\nu}$? I know it is not a definition..but I think that we use $\delta_{\mu\nu}$..

What do you intend $x^2$ to be here? $\delta_{\mu\nu}$ is simply an object you will not see apart from for Cartesian tensors.

 

[stevendaryl]

O.K, I understand $\delta_{\mu\nu}$ is not a tensor, but can't we define $\delta_{\mu\nu}$? For example, can't we write $\partial_\mu \partial_\nu x^2=2\delta_{\mu\nu}$? I know it is not a definition..but I think that we use $\delta_{\mu\nu}$..

You can certainly pick a particular coordinate system, and define a tensor A_{\mu \nu} in that coordinate system to be 1 if \mu = \nu and 0 otherwise. However, it won't have that property in other coordinate systems. So, it would be misleading to call it \delta_{\mu \nu}.

In contrast, g^\mu_\nu has the property that it is 1 if \mu = \nu and 0 otherwise, and this property holds in every coordinate system.

 

[shooride]

What do you intend $x^2$ to be here? $\delta_{\mu\nu}$ is simply an object you will not see apart from for Cartesian tensors.

As far as I understood, the best way to consider $\delta_{\mu\nu}$ is just simply the flat metric.

 

[Orodruin]

As far as I understood, the best way to consider $\delta_{\mu\nu}$ is just simply the flat metric.

This is not correct either, there are several metrics which are flat but not equal to $\delta_{\mu\nu}$, consider all of the metrics in different curvilinear coordinate systems in a Euclidean space. It is also not always possible to chose a flat metric on any given manifold.

 

[robphy]

$\delta^a{}_b$ is probably best thought of as the "index substitution operator" [independent of a metric]
( https://www.google.com/search?q="index+substitution+operator" ).

Given a metric, we can raise and lower indices...
$\delta_a{}_b\equiv\delta^c{}_a g_{cb}=g_{ab}$
...maybe thought of as an operator on a tensor like $A^b$ that does index-lowering, then substitution
...but this reveals itself to be the metric after index-substitution...

Note: $\delta_a{}_b$ is just a convenient abbreviation for $\delta^c{}_a g_{cb}$, although it should probably be abandoned in favor of $g_{ab}$.

 

[vanhees71]

Tip to the OP: Take some book on linear algebra and read about bilinear forms and "Sylvester's Law" of inertia and the signature of fundamental forms (pseudo-metrics or metrics of pseudo-Euclidean and Euclidean vector spaces):

https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia

Then note that in the Ricci calculus it is crucial whether you have upper and lower indices. The place of the index indicates how it transforms under changes of the basis of the vector space and its co-basis.

For differential geometry in pseudo-Riemann spaces and General Relativity you consider tangent and co-tangent spaces. For holonomous bases the transformation laws for tensor components are easy to remember. If you change from coordinates $q^{\mu}$ to $\tilde{q}^{\mu}$ contravariant tensors transform like the increments $\mathrm{d} q^{\mu}$:
$$\mathrm{d} \tilde{q}^{\mu} = \mathrm{d} q^{\nu} \frac{\partial \tilde{q}^{\mu}}{\partial q_{\nu}},$$
i.e., contravariant vector components transform like
$$\tilde{V}^{\mu} = \frac{\partial \tilde{q}^{\mu}}{\partial q_{\nu}} V^{\nu} ={T^{\mu}}_{\nu} V^{\nu}.$$
As you see ${T^{\mu}}_{\nu}$ forms the Jacobian matrix of the coordinate transformation (local diffeommorphism).

Covariant vector components (i.e., components of a one-form) transform as the partial derivatives of a scalar field, i.e., because of
$$\tilde{\partial}_{\mu} \Phi:=\frac{\partial \Phi}{\partial \tilde{q}^{\mu}} = \frac{\partial \Phi}{\partial q^{\nu}} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\mu}},$$
i.e., covariant vector components transform like
$$\tilde{V}_{\mu} = V_{\nu} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\mu}}=V_{\nu} {U^{\nu}}_{\mu}.$$
Now ${U^{\nu}}_{\mu}$ is the inverse Jacobian, because
$${T^{\mu}}_{\nu} {U^{\nu}}_{\rho} = \frac{\partial \tilde{q}^{\mu}}{\partial q^{\nu}} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\rho}} = \frac{\partial \tilde{q}^{\mu}}{\partial \tilde{q}^{\rho}} = {\delta^{\mu}}_{\rho}.$$
The contravariant and covariant vector components transform contragrediently to each other, as it must be, because you want the contraction between a 1-form and a vector, i.e., $V_{\mu} W^{\mu}$ to be a scalar, i.e.,
$$V_{\mu} W^{\mu} = \tilde{V}^{\mu} \tilde{W}_{\mu}.$$
The transformation laws for higher-rank tensor follow now immideately by the definition that the transform like Kronecker products of vectors and 1-forms.